Possible assistance with a monstrous pair of integrals In my study of ODEs I have recently encountered this monster of an integral 
the sum of two integrals:
 $ \int_{0}^{2\pi} \frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx + \int_{0}^{2\pi} \frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx $ 
Even Wolfram is having trouble with this, I was hoping someone could please tell me if I can do it analytically or numerically any way at all? It really is scary looking. Of course $\omega$ and $A$ are parameters with $A$ between 0 and 1.
Edits:
The first one (with the cosine) is zero as the great answers point out but the one with the product of two sines is a mystery.
I feel I owe an explanation: the relation to ODE comes from a previous problem I stated here dealing with the Melnikov integral here the answer changes variables from t to x.
 A: It's simpler than it appears at first :
Let $\quad f(x)=\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}$
$$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx +\int_{\pi}^{2\pi} f(x)dx $$
Let $\quad x=2\pi-t \quad$ It is easy to prove that $\quad f(x)=-f(t)$
because $\sin(x)=\sin(-t)=-\sin(t)$ and $\sin^2\left(\frac{x}{2}\right)=\sin^2\left(\pi-\frac{t}{2}\right)=\sin^2\left(\frac{t}{2}\right)$
$\int_{\pi}^{2\pi} f(x)dx =\int_{\pi}^{0} \left(-f(t)\right)(-dt) =-\int_{0}^{\pi} f(t)dt $
$$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx -\int_{0}^{\pi} f(t)dt=0 $$ 
$$\int_0^{2\pi}\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx=0$$
With the same method and $g(x)=\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx$ 
$g(x)=g(t) \quad\to\quad \int_0^{2\pi}g(x)dx=\int_0^{\pi}g(x)dx+\int_0^{\pi}g(t)dt=2\int_0^{\pi}g(x)dx\neq 0$
$$\int_0^{2\pi}\frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}dx \neq 0$$
In the equation above, the symbol $\neq 0$ means "generally not equal to $0$ ". It doesn't mean "always not equal to $0$ ".
A: Oh, the answer is 0. Note that the function satisfies
$$f(x) = - f(2\pi - x)$$
for any values of $A, \Omega$.
